$x^2+y^2+2axy=0 \Rightarrow x=0$ and $y=0$

Question

Show that for all real numbers $x$ and $y$, for all $-1<a<1$ $$x^2+y^2+2axy=0 \Rightarrow x=0 \text{and} y=0$$ I see that $x^2+y^2+2axy=(ax+y)^2+x^2-(ax)^2$.

I'm stuck here.

Answer 1

Given $$-1 \leq a \leq 1\Rightarrow -2xy \leq 2axy \leq 2xy\Rightarrow x^2+y^2-2xy \leq x^2+y^2+2axy \leq x^2+y^2+2xy$$

So we get $$(x-y)^2\leq x^2+y^2+2axy \leq (x+y)^2$$

Now Given $x^2+y^2+2axy=0$

So we get $(x-y)^2\leq 0$ and $(x+y)^2\geq 0$

So thses two condition are simultaneously hold when $(x-y) = 0$ and $(x+y)=0$

So we get $x=y=0$

Answer 2

Note that $(x+ay)^2=x^2+2axy+a^2y^2$.

\begin{align} (x+ay)^2&\ge0\\ x^2+2axy+a^2y^2&\ge0\\ x^2+2axy&\ge-a^2y^2\\ x^2+2axy+y^2&\ge y^2-a^2y^2\\ x^2+2axy+y^2&\ge y^2(1-a^2) \end{align} Now, if $-1<a<1$, then $a^2<1$ and $0<1-a^2$. Also, unless $y=0$, $y^2$ is always positive. Thus, unless $y=0$, the right-hand-side of the above is positive, and: \begin{align} x^2+2axy+y^2&\ge y^2(1-a^2)>0\\ x^2+2axy+y^2&>0 \end{align} Thus, unless $y=0$, the left-hand-side cannot equal $0$.

If $y$ does equal zero, then $x^2+2axy+y^2=0$ implies $x^2=0$ and thus $x=0$. Thus, if the expression equals zero, then $x=0$ and $y=0$.

Answer 3

$$x=r\cos(\theta)\,\,\,\,,\,\,\,y=r\sin(\theta)\\ \Rightarrow\,\, r^2(1+asin(2\theta))=0$$ if $\,\,\,r^2\neq 0\,\,\,$ we have $\,\,\,1+asin(2\theta)=0\,\,\,$ that is impossible, so $\,\,\,r^2= 0\,\,\,$and we have$\,\,\,x=y=0$.